since it has a diameter of 28, then its radius must be half that or 14.
![\textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=14 \end{cases}\implies A=\pi (14)^2\implies A=196\pi ~\hfill \stackrel{\stackrel{semi-circle}{half~that}}{98\pi }](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20circle%7D%5C%5C%5C%5C%20A%3D%5Cpi%20r%5E2~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D14%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%20%2814%29%5E2%5Cimplies%20A%3D196%5Cpi%20~%5Chfill%20%5Cstackrel%7B%5Cstackrel%7Bsemi-circle%7D%7Bhalf~that%7D%7D%7B98%5Cpi%20%7D)
9000 x .08 = 720
720 x 5= 3600
9000-3600= 5400
5,400
Answer:
The actual area of land is 80 mi² .
Step-by-step explanation:
Given that we have to find the actual area so first we have to square the ratio(scale factor) :
1 in : 5 mi
(1 in)² : (5 mi)²
1 in² : 25 mi²
Next, the actual area will be :
1 in² = 25 mi²
(1 × 3.2) in² = (25 × 3.2) mi²
3.2 in² = 80 mi²